Lines and Angles Test

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Lines and Angles Test - 20

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Question 1
1 / -0

If angle $$P$$ and angle $$Q$$ are supplementary and the measure of angle $$P$$ is $$60^o$$, then the measure of angle $$Q$$ is:

A
$$120^o$$

B
$$60^o$$

C
$$30^o$$

D
$$20^o$$

Solution

Option (a) is correct.
Given that $$\angle P +\angle Q=180^o$$
$$\Rightarrow 60^o +\angle Q=180^o$$
$$\Rightarrow \angle Q=180^o -60^o$$
$$\Rightarrow \angle Q=120^o$$
Question 2
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In the figure, if OP||RS, $$\angle OPQ=110^0 and \angle QRS =130^0$$, then $$\angle PQR$$ is equal to

A
$$40^0$$

B
$$50^0$$

C
$$60^0$$

D
$$70^0$$

Solution

As OP||RS,
$$\angle$$ PQR $$ =180 -(180-110)-(180-130) $$=$$60$$
Question 3
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Lines PQ and RS intersect at O. If $$\angle POS = 2 \angle SOQ$$, then the four angles at O are:

A
$$30^{\circ}, 30^{\circ}, 120^{\circ}, 180^{\circ}$$

B
$$60^{\circ}, 60^{\circ}, 120^{\circ}, 120^{\circ}$$

C
$$60^{\circ}, 90^{\circ}, 90^{\circ}, 120^{\circ}$$

D
$$30^{\circ}, 60^{\circ}, 90^{\circ}, 180^{\circ}$$

Solution

PQ and RS intersect at O. then,
$$\angle POS = \angle QOR$$ (vertically opposite angles)
$$\angle SOQ =\angle POR$$ (vertically opposite angles)
Given $$\angle POS = 2\angle SOQ$$
Sum of all angles = 360
$$\angle POS + \angle SOQ + \angle QOR +\angle ROP = 360$$
$$6\angle SOQ = 360$$
$$\angle SOQ = 60$$
Hence, the four angles = $$60^o, 60^0, 120^0, 120^0$$.
Question 4
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If one of the angles formed by two intersecting lines is a right angle, what can you say about the other three angles?

A
$$45^{\circ}, 45^{\circ}, 180^{\circ}$$

B
$$90^{\circ}, 90^{\circ}, 90^{\circ}$$

C
$$60^{\circ}, 60^{\circ}, 90^{\circ}$$

D
$$60^{\circ}, 90^{\circ}, 90^{\circ}$$

Solution

Let $$AB$$ and $$CD$$ be two lines Intersecting at $$O$$, such that, $$\angle AOD = 90^{\circ}$$

Now, $$\angle AOD = \angle COB = 90^{\circ}$$ (Vertically opposite angles)
$$\Longrightarrow \angle AOD + \angle DOB = 180^o$$ (Angles on a straight line)
$$\Longrightarrow 90 + \angle DOB = 180^o$$
$$\angle DOB = 90^{\circ}$$

$$\angle DOB = \angle AOC = 90^{\circ}$$ (Vertically opposite angles)
Thus, all angles are $$90^{\circ}$$.
Question 5
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What value of $$\angle$$FEC will make $$CD \parallel EF$$, and AB||CD ?

A
150$$^{\circ}$$

B
145$$^{\circ}$$

C
140$$^{\circ}$$

D
135$$^{\circ}$$

Solution

Consider AB || CD || EF
then , $$\angle ABC = \angle BCD = 65^{\circ}$$ (corresponding angles)
Therefore, $$\angle BCE + \angle ECD = 65^0$$
$$30 + \angle ECD = 65^0$$
$$\angle ECD = 35^{\circ}$$
Now, $$\angle FEC + \angle ECD = 180^{\circ}$$ (supplementary angles)
$$ 35 + \angle FEC = 180$$
$$\angle FEC = 145^{\circ}$$.
Question 6
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From the given figure, we can say that

A
$$l \parallel m $$, $$p \parallel q$$

B
$$l \parallel m$$, $$p \not\parallel q$$

C
$$l \not\parallel m$$, $$p \parallel q$$

D
$$l \not\parallel m$$, $$p \not\parallel q$$

Solution

Sum of angle between the lines $$l, m$$
$$32 + 48 = 80^{\circ}$$
Since, the sum is not $$180^{\circ}$$, hence, $$l$$ is not parallel to $$m$$.

Sum of angles between the lines $$p, q$$
$$73 + 106 = 179^{\circ}$$
The sum of angles between two lines is less than $$180^{\circ}$$, thus $$p$$ is not parallel to $$q$$
Question 7
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In the given figure, $$l||m||n$$, If $$x : y = 5 : 4$$, then the measure of angle z is :

A
$$40^{\circ}$$

B
$$50^{\circ}$$

C
$$90^{\circ}$$

D
$$80^{\circ}$$

Solution

Let $$x=5a$$ and $$y=4a$$

Now $$\angle y+\angle 1=180^{\circ}$$ (Angle made on straight line $$(n)$$ are supplementary)

$$\Rightarrow 4a+\angle 1={ 180 }^{ \circ }\\ \Rightarrow \angle 1={ 180 }^{ \circ }-4a..........(i)$$

Now $$m||n$$ and $$t$$ is transversal

$$\therefore \angle x=\angle 1$$ (Corresponding angles)

$$\Rightarrow 5a=\angle 1\\ \Rightarrow 5a={ 180 }^{ \circ }-4a\\ \Rightarrow 9a={ 180 }^{ \circ }\\ \Rightarrow a={ 20 }^{ \circ }$$

Now $$\angle y=4a=4\times { 20 }^{ \circ }={ 80 }^{ \circ }$$

Now $$l||n$$ and $$t$$ is transversal.

$$\therefore \angle z=\angle y$$ (Alternate angles)

$$\Rightarrow \angle z={ 80 }^{ \circ }$$
Question 8
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The angle which exceeds its complement by $$20^{\circ}$$ is:

A
$$45^{\circ}$$

B
$$55^{\circ}$$

C
$$70^{\circ}$$

D
$$110^{\circ}$$

Solution

Let the required angle be $$x$$.
We knows, complementary angles $$=$$ Sum of two angles is $$90^o$$.
$$\therefore x = (90^o - x ) + 20^o $$
$$\therefore$$ $$x = 90^o - x + 20^o$$
$$\therefore$$ $$ 2x = 110^o$$
$$\therefore x = 55^o$$.
Hence, option $$B$$ is correct.
Question 9
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In the given figure, $$\angle QPB$$ is,

A
$$60^{\circ}$$

B
$$45^{\circ}$$

C
$$30^{\circ}$$

D
$$15^{\circ}$$

Solution

In the given figure,
$$\angle APR + \angle RPQ + \angle QPB = 180^{\circ}$$
$$2x + 3x+x = 180^{\circ}$$
$$6x = 180^{\circ}$$
$$x = 30^{\circ}$$
Question 10
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In the given figure, $$ AE\parallel BD $$, $$ AC\parallel ED $$ and $$ AB = AC $$. Find $$ \angle a $$.

A
$$63^o$$

B
$$64^o$$

C
$$62^o$$

D
None of the above

Solution

Here, $$\angle EDC = 58^{\circ}$$ (Vertically opposite angles)

Given $$AE \parallel CD$$ so,
$$\implies$$ $$\angle c = \angle EDC = 58^{\circ}$$ [Corresponding angles]

Also, $$AC \parallel ED$$
$$\implies$$ $$\angle ACB =\angle EDC= 58^{\circ}$$ [Corresponding angles)

Since, $$AB = AC$$
$$\implies$$ $$\angle ABC = \angle ACB=58^o$$ [Isosceles triangle property]

Now, in $$\triangle ABC$$,
$$\angle ABC + \angle ACB + \angle BAC = 180^o $$
$$\implies$$ $$58^o + 58^o + \angle a = 180^o$$
$$\implies$$ $$116^o + \angle a = 180^o$$
$$\implies$$ $$\angle a = 180^o - 116^o$$
$$\implies$$ $$\angle a = 64^{o}$$

Hence, option $$B$$ is correct.